find the coordinate of the foot of perpendicular form of point (-1,3)to the line(3x-4y-16)=0 - Maths - Straight Lines

Question

find the coordinate of the foot of perpendicular form of point
(-1,3)to the line(3x-4y-16)=0

Kirthiga Suresh · Accepted Answer

Find the coordinate of the foot of perpendicular from point $(- 1, 3)$ to the line $3 x - 4 y - 16 = 0$ .

Solution:

The line $3 x - 4 y - 16 = 0$ can be rearranged as:

$4 y = 3 x - 16$

$y = \frac{3}{4} x - 4$

Comparing this with standard equation of a straight line $y = m x + c$ where $m$ is the slope of the line and $c$ is the intercept on y-axis, slope the line is equal to $\frac{3}{4}$ .

Thus, slope of the line perpendicular to the given line would be: $- \frac{1}{m}$ , i.e, $- \frac{1}{\frac{3}{4}}$

that is, $- \frac{4}{3}$

Equation of the perpendicular line would be of the form:

$y = m x + c$ where $m = - \frac{4}{3}$

Thus, $y = - \frac{4}{3} x + c$ ................(1)

is the equation of the line perpendicular to line $3 x - 4 y - 16 = 0$

As the line passes thru point $(- 1, 3)$ ,

Substituting $x = - 1, y = 3$ in (1),

$3 = (- \frac{4}{3}) \times (- 1) + c$

$c = 3 - \frac{4}{3} = \frac{5}{3}$

Thus equation of the line perpendicular to the line $3 x - 4 y - 16 = 0$ from point $(- 1, 3)$ is:

$y = (- \frac{4 x}{3} + \frac{5}{3})$

$3 y = - 4 x + 5$

$4 x + 3 y - 5 = 0$